In the xy-coordinate plane, the graph of the equation y=2x^2-12x-32 has zeros at x=d and x=e, where d>e. the graph has a minimum at (f, -50). What are the values of d, e, and f?
To find the values of d, e, and f, we first need to find the x-coordinates of the zeros and the minimum point.
1. Zeros of the equation:
Set y = 0 and solve for x:
0 = 2x^2 - 12x - 32
0 = x^2 - 6x - 16
0 = (x + 2)(x - 8)
So, the zeros are at x = -2 and x = 8. Since d>e, d = 8 and e = -2.
2. Minimum point:
To find the x-coordinate of the minimum point, we use the formula for the x-coordinate of the vertex of a parabola:
x = -b / 2a
x = -(-12) / (2*2)
x = 12 / 4
x = 3
Therefore, the minimum point occurs at x = 3. Since the minimum point is (f, -50), f = 3.
So, the values of d, e, and f are:
d = 8
e = -2
f = 3